2
$\begingroup$

I have questions about the proof of Theorem 14 (poisson's formula for half-space) in Page 38.

Let $K(x,y)$ be the Poisson's kernel for $\mathbb R^n_+$: $$K(x,y)=\frac{2x_n}{na(n)}\frac{1}{|x-y|^n}dy\quad (x\in\mathbb R^n_+,\; y\in \partial \mathbb R_+^n).$$

Let $$u(x)=\frac{2x_n}{na(n)}\int_{\partial \mathbb R_+^n}\frac{g(y)}{|x-y|^n}dy\quad (x\in \mathbb R_+^n).$$

The author claims that since $x\mapsto K(x,y)$ is smooth for $x\neq y$, we easily verify as well $u\in C^\infty(R_+^n)$, with $$\Delta u(x)=\int_{\partial\mathbb R_+^n}\Delta_xK(x,y)g(y)dy=0\quad (x\in \mathbb R_+^n).$$ Suppose $g\in C(R^{n-1})\cap L^\infty(R^{n-1})$. $g$ is bounded.

My questions are

  1. How to prove $u\in C^\infty(\mathbb R_+^n)$?
  2. How to prove that we can take the laplacian operator into the integration when calculating $$\Delta u(x)=\Delta_x\int_{\partial \mathbb R_+^n}K(x,y)g(y)dy\;?$$

Thanks!

$\endgroup$
4
  • $\begingroup$ What is the assumption on $g$? $\endgroup$
    – Martin
    Jul 13, 2016 at 21:14
  • $\begingroup$ @Martin $g\in C(R^{n-1})\cap L^\infty(R^{n-1})$. $g$ is bounded. $\endgroup$ Jul 13, 2016 at 21:18
  • $\begingroup$ Convolutions tend to give the same smoothness as the smoothest function. Since $g(x) \in C^{\infty}$ you should get $u \in C^\infty$. Now all the estimates on the limit of finite differences (in order to pass the limit through) are easier if you have compact support for $g$. I would add this assumption, and see if there is some way you can uniformly approximate $g$ as compactly supported functions. Look up mollifiers on wikipedia or elsewhere to get some clues in how to proceed in these types of problems. $\endgroup$
    – abnry
    Jul 13, 2016 at 21:37
  • $\begingroup$ @nayrb Thanks. Can you write a detailed argument? $\endgroup$ Jul 13, 2016 at 23:33

1 Answer 1

0
$\begingroup$

This should prove that you can differentiate under the integral: $$\partial_i u(x) = \lim_{h \to 0}\frac{u(x+e_i h)-u(x)}{h} = \lim_{h \to 0}\frac{\int_{\partial R_+^n} K(x+e_i h,y)g(y)dy -\int_{\partial R_+^n} K(x,y)g(y)dy}{h} = \lim_{h \to 0} \int_{\partial R_+^n} \frac{K(x+e_i h,y) -K(x,y)}{h}g(y)dy = \int_{\partial R_+^n} \partial_{i,x}K(x,y)g(y)dy$$

(Passing the last limit inside uses dominated convergence). Smoothness of $u$ now follows from smoothness of $K$ and $g$.

EDIT: On the dominated convergence. Since $K$ is smooth we can apply the mean value theorem to estimate $\frac{K(x+e_i h,y) -K(x,y)}{h}$, for $|h|< \frac{x_n}{2}$ (just making sure that we do not hit $\partial R_+^n$). Thus for bounded values of $y$ say $|y-x'|< 1$ ($x'=(x_1,\ldots,x_{n-1})$) we may bound the integrand by a constant. Outside this ball we see that the derivative of the kernel wrt. $x_i$ is less than the function, which is constructed from the partial derivative by replacing $x$ with $(x',0)$. Outside $|y-x'|<1$ we now have a function that decreases like $|y-x'|^{-(n+1)}$, whereas the radial volume element goes like $r^{n-1}dr$. Hence the outside integral remains finite, and we have constructed a $L^1$ bound, and we may now use the dominated convergence to pass the limit inside.

$\endgroup$
10
  • $\begingroup$ thanks for your answer. Can you write it more clear about how to use dominated convergence theorem here? Thanks! $\endgroup$ Jul 13, 2016 at 21:45
  • $\begingroup$ It turned out that the answer to that question is a bit more technical than expected, and I'll have to get back to you on that. The main idea is that near the projection of $x$ onto $R_+^{n-1}$ $x'$ we may bound $K$ by a constant, and far from $x'$ the kernel decreases fast enough to have a finite integral. Thus we can construct a dominating $L^1$ function. But as i stated the details are more technical than I expected. $\endgroup$
    – Martin
    Jul 13, 2016 at 22:05
  • $\begingroup$ Thanks! Can you also give a detailed explanation about how to derive the smoothness of $u$ from the smoothness of $K$ and $g$ latter? $\endgroup$ Jul 13, 2016 at 22:22
  • $\begingroup$ Note that when you derive the kernel wrt. $x$ it gives a function, that decreases at least as fast as $K$ for large $y$. Thus you can repeat the above argument for any partial derivative. $\endgroup$
    – Martin
    Jul 14, 2016 at 11:20
  • $\begingroup$ In the the domain of $|y-x'|<1$, how to show that the integrand is bounded? I notice that $K(x,y)$ may blow up when $y\rightarrow x$. $\endgroup$ Jul 14, 2016 at 16:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .